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%%文档的题目、作者与日期
\author{学号 \underline{\hspace{4cm}} \hspace{1cm} 姓名 \underline{\hspace{4cm}} }
\title{复变函数测验1}
%\date{\vspace{-3ex}}
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\date{2024 年 5 月 6 日}
%\date{March 9, 2021}

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\begin{document}

\maketitle

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\begin{enumerate}

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\item  %Problem 1
设三个复数 $z_1,z_2,z_3$ 成为一个等腰直角三角形的顶点。写出这三个复数满足的等式。

\vspace{0.0cm}

%{\color{red}解答：设 $z_1$ 是直角顶点。则 $z_2-z_1$ 旋转 90度成为 $z_3-z_1$. 因此有
%$$(z_3-z_1) = \pm i (z_2-z_1). $$
%平方可得 
%$$(z_3-z_1)^2 + (z_2-z_1)^2 = 0. $$
%
%}

\vspace{4.0cm}

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\item  %Problem 2
分别设 $z=x+iy$ 和 $z=re^{i\theta}$, 计算函数 $w=z^3+3z+1$ 的表达式。

\vspace{0.0cm}

%{\color{red}解答：设 $z=x+iy$, 则 
%\begin{eqnarray*}
%w &=& (x+iy)^3+3(x+iy)+1 \\ 
%&=& x^3+3x^2(iy)+3x(iy)^2+(iy)^3+3(x+iy)+1 \\ 
%&=& (x^3-3xy^2+3x+1) + i(3x^2y-y^3+3y). 
%\end{eqnarray*}
%
%设 $z=re^{i\theta}$, 则
%\begin{eqnarray*}
%w &=& (re^{i\theta})^3 + 3(re^{i\theta}) +1 \\ 
%&=& r^3e^{i3\theta} + 3re^{i\theta} +1 \\ 
%&=& (r^3\cos(3\theta)+3r\cos\theta+1) + i(r^3\sin(3\theta)+3r\sin\theta).
%\end{eqnarray*}
%
%}

\vspace{5.0cm}

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\item  %Problem 3
函数 $w=\frac{1}{z}$ 将 $z$ 平面上的曲线 $x^2+(y-1)^2=1$ 变成 $w$ 平面上的什么曲线？

\vspace{0.0cm}

%{\color{red}解答：从 $z=\frac{1}{w}$ 可得 
%$$x+iy = \frac{1}{u+iv} = \frac{u-iv}{u^2+v^2}. $$
%因此有
%$$x=\frac{u}{u^2+v^2},\,\,\, y=\frac{-v}{u^2+v^2}.$$
%代入 $x^2+(y-1)^2=1$ 得到关于 $u,v$ 的等式为
%$$\frac{u^2}{(u^2+v^2)^2} + \left( \frac{-v}{u^2+v^2}-1 \right)^2=1.$$
%两边同乘以 $(u^2+v^2)^2$, 得到 
%$$u^2 + (-v- u^2-v^2)^2 = (u^2+v^2)^2.$$
%第二项平方展开得到
%$$u^2+v^2+(u^2+v^2)^2+2v(u^2+v^2) = (u^2+v^2)^2.$$
%消去可得 $1+2v=0$. 即 $v=-\frac{1}{2}$. 
%}

\vspace{0.0cm}

\newpage

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\item  %Problem 4
设 $|z|=1$, 证明 $$\left\vert \frac{az+b}{\bar{b}z+\bar{a}} \right\vert =1. $$

\vspace{0.0cm}

%{\color{red}解答：设 $w=\frac{az+b}{\bar{b}z+\bar{a}}$, 现计算 $w\bar{w}$. 
%\begin{eqnarray*}
%w\bar{w} &=& \left(\frac{az+b}{\bar{b}z+\bar{a}}\right) \left(\frac{\bar{a}\bar{z}+\bar{b}}{b\bar{z}+a} \right) \\ 
%&=& \frac{(az+b)(\bar{a}\bar{z}+\bar{b})}{(\bar{b}z+\bar{a})(b\bar{z}+a)} \\ 
%&=& \frac{ a\bar{a}z\bar{z} + b\bar{b} + a\bar{b}z + \bar{a}b\bar{z} }{ b\bar{b}z\bar{z} + a\bar{a} + a\bar{b}z + \bar{a}b\bar{z} }
%\end{eqnarray*}
%代入 $z\bar{z}=1$, 可得
%\begin{eqnarray*}
%w\bar{w} &=& \frac{ a\bar{a} + b\bar{b} + a\bar{b}z + \bar{a}b\bar{z} }{ b\bar{b} + a\bar{a} + a\bar{b}z + \bar{a}b\bar{z} } =1. 
%\end{eqnarray*}
%这个结论表明函数 $f(z)=\frac{az+b}{\bar{b}z+\bar{a}}$ 将单位圆变到单位圆。
%}

\vspace{4.0cm}

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\item  %Problem 5
已知平行四边形的三个顶点对应的复数为 $z_1=-1+2i$, $z_2=3-i$ 和 $z_3=2+i$, 求第四个顶点对应的复数。

\vspace{0.0cm}

%{\color{red}解答：设第四个顶点为 $z_4$. 将原点平移到 $z_2$, 根据复数加法的平行四边形法则，可得
%$$(z_1-z_2) + (z_3-z_2) = z_4-z_2. $$
%因此可以求出 $z_4$ 为
%\begin{eqnarray*}
%z_4 &=& z_1 + z_3 - z_2 \\
% &=& (-1+2i) + (2+i) - (3-i) \\ 
% &=& -2 +4i.
%\end{eqnarray*}
%
%
%}

\vspace{4.0cm}


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\item  %Problem 6
证明函数 $f(z) = x^2-y^2+2x+2xyi+2yi+3$ 在复平面上解析，并求其导数。

\vspace{0.0cm}

%{\color{red}解答：记 $f(z)=u+iv$, 则有 
%\begin{eqnarray*}
%\left\{\begin{array}{rcl}
%u &=& x^2-y^2+2x+3, \\
%v &=& 2xy+2y.
%\end{array}\right.
%\end{eqnarray*}
%计算偏导数可得 $u_x = 2x+2, u_y=-2y, v_x = 2y, v_y=2x+2$.
%可见四个偏导数在复平面上处处连续，且柯西黎曼方程成立。
%根据定理2.5可得函数 $f(z)$ 在复平面上解析，其导数为
%$$
%f'(z)=u_x+iv_x = 2x+2+2yi. 
%$$
%
%}

\vspace{4.0cm}


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\item  %Problem 7
求 $\sin(3i)$ 和 $\cos(3i)$ 的值，并验证 $\sin^2(3i) + \cos^2(3i) =1$. 

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%{\color{red}解答：根据正弦函数和余弦函数的定义，可得
%\begin{eqnarray*}
%\sin(3i) &=& \frac{e^{i(3i)} - e^{-i(3i)}}{2i} = \frac{ e^{-3} - e^{3} }{2i} = \frac{e^3-e^{-3}}{2}i, \\  
%\cos(3i) &=& \frac{e^{i(3i)} + e^{-i(3i)}}{2} = \frac{ e^{3} + e^{-3} }{2}. 
%\end{eqnarray*}
%代入计算平方和，可得
%\begin{eqnarray*}
%\sin^2(3i) + \cos^2(3i) = \left(\frac{e^3-e^{-3}}{2}i \right)^2 + \left( \frac{ e^{3} + e^{-3} }{2} \right)^2 = 
%\frac{e^6 + e^{-6}-2}{4}(-1) + \frac{e^6+e^{-6}+2}{4} = 1. 
%\end{eqnarray*}
%
%}

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\end{enumerate}


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\end{document}

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